infinite

Haskell infinite recursion in list comprehension

依然范特西╮ 提交于 2019-11-28 13:07:38
I am trying to define a function that accepts a point (x,y) as input, and returns an infinite list corresponding to recursively calling P = (u^2 − v^2 + x, 2uv + y) The initial values of u and v are both 0. The first call would be P = (0^2 - 0^2 + 1, 2(0)(0) + 2) = (1,2) Then that resulting tuple (1,2) would be the next values for u and v, so then it would be P = (1^2 - 2^2 + 1, 2(1)(2) + 2) = (-2,6) and so on. I'm trying to figure out how to code this in Haskell. This is what I have so far: o :: Num a =>(a,a) -> [(a,a)] o (x,y) = [(a,b)| (a,b)<- [p(x,y)(x,y)]] where p(x,y)(u,v) = ((u^2)-(v^2)

上拉加载底部转圈圈动画并文字提示

与世无争的帅哥 提交于 2019-11-28 12:33:09
<block> <view class="listFooterLoading" wx:if="{{loadStatus==1}}"> <!-- 转圈圈 --> <view class="m-load2"> <view class="line"><view></view><view></view><view></view><view></view><view></view><view></view></view> <view class="circlebg"></view> </view> <view class='load-word'>加载中</view> </view> <text class="listFooter" wx:elif="{{loadStatus==2}}">没有更多数据了!</text> <text class="listFooter" wx:elif="{{loadStatus==-1}}">数据加载失败,请检查网络</text> <text class="listFooter" wx:elif="{{loadStatus==-2}}" >没有任何数据!</text> </block> .listFooter{ color: #888888; background: #EFEFF4; display: block; text-align: center;

cuda infinite kernel

穿精又带淫゛_ 提交于 2019-11-28 09:29:35
问题 I am working on an application for which it is necessary to run a CUDA kernel indefinitely. I have one CPU thread that writes stg on a list and gpu reads that list and resets (at least for start). When I write inside the kernel while(true) { //kernel code } the system hangs up. I know that the GPU is still processing but nothing happens of course. And I am not sure that the reset at the list happens. I have to mention that the GPU used for calculations is not used for display, so no watchdog

Is there a standard class for an infinitely nested defaultdict?

瘦欲@ 提交于 2019-11-28 08:33:12
Does anyone know if there's a standard class for an infinitely nestable dictionary in Python? I'm finding myself repeating this pattern: d = defaultdict(lambda: defaultdict(lambda: defaultdict(int))) d['abc']['def']['xyz'] += 1 If I want to add "another layer" (e.g. d['abc']['def']['xyz']['wrt'] ), I have to define another nesting of defaultdicts. To generalize this pattern, I've written a simple class that overrides __getitem__ to automatically create the next nested dictionary. e.g. d = InfiniteDict(('count',0),('total',0)) d['abc']['def']['xyz'].count += 0.24 d['abc']['def']['xyz'].total +=

Android: How can I stop an infinite animation applied on an ImageView?

♀尐吖头ヾ 提交于 2019-11-28 08:06:33
I have an ImageView on which I have applied a rotate animation. Since I want the rotation to go on continuously, I gave the repeatCount as infinite in my rotate.xml: android:repeatCount="infinite" In onCreate(), I load the animation and start it. Animation myAnim = AnimationUtils.loadAnimation(this, R.anim.rotate); objectImg.startAnimation(myAnim); When a button is pressed, the rotation must stop. Hence in my onClick(), I called clearAnimation(). objectImg.startAnimation(myAnim); My simple question is whether stopping the animation is the right thing to do. I assume clearAnimation()

吸引注意力的动画

别来无恙 提交于 2019-11-28 06:24:09
<p><strong>单击</strong> 按钮以激活其动画。</p> <div class='btn-container'> <button class='btn btn--shockwave is-active'> Shockwave </button> <button class='btn btn--jump'> Jump </button> <button class='btn btn--pulse'> Pulse </button> <button class='btn btn--blink'> Blink </button> <button class='btn btn--wiggle'> Wiggle </button> <button class='btn btn--wut'> Hatching Alien </button> </div>    body { max-width: 600px; margin: 4.8rem auto; background: #acd4e4; font-family: tahoma, helvetica neue, helvetica, arial, sans-serif; background: linear-gradient(to right, #acd4e4, #bbe0df); text-align: center; }

What's exactly happening in infinite nested lists?

让人想犯罪 __ 提交于 2019-11-28 04:55:35
It's possible to create an infinite nested list in Python. That's clear and, although not popular and definitely not useful is a known fact. >>> a = [0] >>> a[0] = a >>> a [[...]] >>> a[0] == a True My question is, what is happening here: >>> a = [0] >>> b = [0] >>> a[0], b[0] = b, a >>> a [[[...]]] >>> b [[[...]]] >>> a == b Traceback (most recent call last): File "<stdin>", line 1, in <module> RuntimeError: maximum recursion depth exceeded in cmp >>> a[0] == b True >>> a[0][0] == b Traceback (most recent call last): File "<stdin>", line 1, in <module> RuntimeError: maximum recursion depth

Left and Right Folding over an Infinite list

ⅰ亾dé卋堺 提交于 2019-11-28 03:32:06
I have issues with the following passage from Learn You A Haskell (Great book imo, not dissing it): One big difference is that right folds work on infinite lists, whereas left ones don't! To put it plainly, if you take an infinite list at some point and you fold it up from the right, you'll eventually reach the beginning of the list. However, if you take an infinite list at a point and you try to fold it up from the left, you'll never reach an end! I just don't get this. If you take an infinite list and try to fold it up from the right then you'll have to start at the point at infinity, which

R - svd() function - infinite or missing values in 'x'

我们两清 提交于 2019-11-28 00:25:52
I am constantly getting this error. I am sure the matrix does not have any non-numeric entries. I also tried imputing the matrix, did not work. Anyone know what the error might be? fileUrl <- "https://dl.dropboxusercontent.com/u/76668273/kdd.csv"; download.file(fileUrl,destfile="./kdd.csv",method="curl"); kddtrain <- read.csv("kdd.csv"); kddnumeric <- kddtrain[,sapply(kddtrain,is.numeric)]; kddmatrix <- as.matrix(kddnumeric); svd1 <- svd(scale(kddmatrix)); You have columns composed of all zeroes. Using scale on a column of all zeroes returns a column composed of NaN . To solve this, remove

How to tell if a list is infinite?

▼魔方 西西 提交于 2019-11-27 21:06:41
问题 Is there a way to tell if a list in Haskell is infinite? The reason is that I don't want to apply functions such as length to infinite lists. 回答1: Applying length to unknown lists is generally a bad idea, both practically due to infinite lists, and conceptually because often it turns out that you don't actually care about the length anyway. You said in a comment: I'm very new to Haskell, so now, don't infinite structures make my programs very vulnerable? Not really. While some of us wish